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Solved Problem on Dynamics

In the system of the figure, the body *B* slides on a horizontal surface without friction, it is connected through
by ropes and pulleys, lightweight and frictionless, with two bodies *A* and *C* that move vertically. The
masses of *A*, *B*, and *C* are, respectively, 5 kg, 2 kg and 3 kg. Find the acceleration of the system
and the tension in the cord. Assume
*g* = 10 m/s^{2}.

Problem data

- mass of body
*A*:*m*_{A}= 5 kg; - mass of body
*B*:*m*_{B}= 2 kg; - mass of body
*C*:*m*_{C}= 3 kg; - free-fall acceleration:
*g*= 10 m/s^{2}.

Problem diagram

We choose the acceleration in the direction in which the body*A* moves downward.
Drawing a free-body diagram for each block and using *Newton's Second Law*

We choose the acceleration in the direction in which the body

\[ \bbox[#99CCFF,10px]
{\vec{F}=m\vec{a}}
\]

Solution

Body*A*
*Newton's Second Law*
we have

Body

- \( \vec W_{\text{A}} \): weight of body
*A*; - \( \vec T_{\text{AB}} \): tension in the cord between blocks
*A*and*B*.

figure 2

\[
P_{\text{A}}-T_{\text{AB}}=m_{\text{A}}a \tag{I}
\]

Body *B*

Vertical direction

Vertical direction

- \( \vec W_{\text{B}} \): weight of body
*B*; - \( \vec N_{\text{B}} \): normal force of the surface on the body.

- \( \vec T_{\text{AB}} \): tension in the rope between blocks
*A*and*B*; - \( \vec T_{\text{BC}} \): tension in the rope between blocks
*B*and*C*.

figure 3

In the vertical direction, the weight and the normal force cancel out, there is no vertical motion.

In the horizontal direction by using*Newton's Second Law* we have

In the horizontal direction by using

\[
T_{\text{AB}}-T_{\text{BC}}=m_{\text{B}}a \tag{II}
\]

Corpo *C*
*Newton's 2nd Law*
we have

- \( \vec W_{\text{C}} \): weight of body
*C*; - \( \vec T_{\text{BC}} \): tension in the rope between blocks
*B*and*C*.

figure 4

\[
T_{\text{BC}}-W_{\text{C}}=m_{\text{C}}a \tag{III}
\]

With equations (I), (II) and (III) we have a system of linear equations with three variables (
\[
\frac{
\left\{
\begin{array}{rr}
W_{\text{A}}-\cancel{T_{\text{AB}}}&=m_{\text{A}}a\\
\cancel{T_{\text{AB}}}-\cancel{T_{\text{BC}}}&=m_{\text{B}}a\\
\cancel{T_{\text{BC}}}-W_{\text{C}}&=m_{\text{C}}a
\end{array}
\right.}
{W_{\text{A}}-W_{\text{C}}=\left(m_{\text{A}}+m_{\text{B}}+m_{\text{C}}\right)a}
\]

\[
a=\frac{W_{\text{A}}-W_{\text{C}}}{m_{\text{A}}+m_{\text{B}}+m_{\text{C}}} \tag{IV}
\]

The weight of bodies
\[
W_{\text{A}}=m_{\text{A}}g\ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ W_{\text{C}}=m_{\text{C}}g \tag{V}
\]

substituting expression (V) in (IV) and the values given in the problem
\[
a=\frac{m_{\text{A}}g-m_{\text{C}}g}{m_{\text{A}}+m_{\text{B}}+m_{\text{C}}}\\
a=\frac{5 \times 10-3 \times 10}{5+2+3}\\
a=\frac{50-30}{10}\\
a=\frac{20}{10}
\]

\[ \bbox[#FFCCCC,10px]
{a=2\;\text{m/s}^{2}}
\]

Substituting the mass of the body
\[
m_{\text{A}}g-T_{\text{AB}}=m_{\text{A}}a\\
5 \times 10-T_{\text{AB}}=5 \times 2\\
50-T_{\text{AB}}=10\\
T_{\text{AB}}=50-10
\]

\[ \bbox[#FFCCCC,10px]
{T_{\text{AB}}=40\ \text{N}}
\]

Substituting the mass of the body
\[
T_{\text{BC}}-m_{\text{C}}g=m_{\text{C}}a\\
T_{\text{BC}}-3.10=3.2\\
T_{\text{BC}}-30=6\\
T_{\text{BC}}=6+30
\]

\[ \bbox[#FFCCCC,10px]
{T_{\text{AB}}=36\ \text{N}}
\]

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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .